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%I #16 Dec 09 2024 23:23:29
%S 1,1,1,1,3,1,3,1,1,1,3,17,1,5,9,1,3,5,3,1,3,5,3,1,3,1,1,17,3,41,1,1,3,
%T 5,129,1,1,5,1,1,3,5,3,17,9,5,3,1,3,1,9,1,3,5,3,1,3,5,3,17,3,1,3,1,33,
%U 5,3,1,9,115825,3,97,3,5,99,1,129,5,3,1,1,5,3,17,11,5,3,1,3,35425,1,1
%N a(n) = gcd(binomial(2n,n), 2^n + 1).
%H Harry J. Smith, <a href="/A066975/b066975.txt">Table of n, a(n) for n = 1..1000</a>
%t Table[GCD[Binomial[2n,n],2^n+1],{n,100}] (* _Harvey P. Dale_, Aug 26 2012 *)
%o (PARI) a(n) = { gcd(binomial(2*n, n), 2^n + 1) } \\ _Harry J. Smith_, Apr 12 2010
%K nonn
%O 1,5
%A _Benoit Cloitre_, Jan 26 2002